VVT15 - Constant temperature gradient for a distorted mesh

Solution Geometry Test case
Finite volume method Case A SVTEST19
Case B SVTEST20
Finite element method Case A SVTEST245
Case B SVTEST246

Description

The purpose of this test case is to verify that the computed element temperatures closely match the theoretical results using the element center conduction method for a flat plate (A) meshed with 3-node planar elements and plate (B) meshed with 4-node planar elements.

Geometry

The geometries for cases (A) and (B) are defined using 1 m square plates with 0.5 m diameter circles in the center and bounded planes. Both cases (A) and (B) are defined as 1 mm thick. Case (A) geometry consists of two bounded planes: one plane is 1 m2 less a circular area 0.5 m in diameter in the center; the second plane in case (A) is a circular area 0.5 mm in diameter located in the center of the 1 m2 plane. Case (B) geometry consists of 5 bounded planes: four planes make up the four quadrants which, when combined, make an area 1 m2 less a circular area 0.5 m in diameter in the center; the second plane in case (B) is a circular area 0.5 m in diameter located in the center of the 1 m2 plane.

Simulation model

The finite element meshes are created using 3-node planar elements (triangles) for case A and 4-node planar elements (rectangles) for case B. The following figures show the mesh for case A and case B.

The meshed elements have the following material and physical properties:

  • Thermal conductivity: k = 200 W/m·°C
  • Mass density: ρ = 1 kg/m3
  • Thickness: 1 mm

The following boundary conditions are applied:

  • Temperature constraint on the top of the plate with a value of T top = 100 °C.
  • Temperature constraint on the bottom of the plate with a value of Tbottom = 0 °C.

The following solution options are set:

  • Solution Type = Steady State
  • Results Options: Select heat fluxes and temperature at nodes and elements.

The default solver parameters are selected.

Theory

For one-dimensional, steady-state conduction in a plane wall with no heat generation and constant thermal conductivity, the temperature varies linearly with x only. According to that, lines of constant temperature must be parallel to surfaces of the plane wall. The temperature distribution is given by [6] :

Results

The following tables compare the computed nodal temperatures with theoretical values. The x location for each case is determined by the x coordinate of the nodes.

Temperatures for case A

Finite volume method Finite element method
Parameter x (m) Ttheory (ºC) Tsim (ºC) Error (%) Tsim (ºC) Error (%)
0 0 0 0 0 0
0.13 13.028 13.027 0.01 13 0.215
0.25 25 25 0.00 25 0
0.364 36.410 36.417 0.02 36.4 0.027
0.486 48.57 48.571 0.00 48.6 0.062
0.558 55.83 55.829 0.00 55.8 0.054
0.65 64.96 64.963 0.00 65 0.062
0.75 75 75 0.00 75 0
0.87 86.98 86.97 0.01 87 0.023
1 100 100 0.00 100 0

Temperatures for case B

Finite volume method Finite element method
Parameter x (m) Ttheory (ºC) Tsim (ºC) Error (%) Tsim (ºC) Error (%)
0.00 0.00 0.00 0.00 0.00 0.00
0.081 8.105 8.106 0.01 8.1 0.062
0.1668 16.68 16.68 0.00 16.68 0.00
0.2585 25.85 25.852 0.00 25.85 0.00
0.365 36.502 36.502 0.00 36.5 0.005
0.5 50 50 0.00 50 0.00
0.5589 55.892 55.893 0.00 55.89 0.004
0.667 66.66 66.66 0.00 66.7 0.060
0.7415 74.148 74.148 0.00 74.15 0.003
0.8338 83.378 83.378 0.00 83.38 0.002
0.92 91.984 91984 0.00 92 0.017
1.00 100.00 100.00 0.00 100 0